1.1. Principle of Working of Barometers

So, this is the barometer. This is the principle of working of barometers. Now, a simple question is, what is the local atmospheric pressure when R is 750 millimeters of Hg? This is R. We are just going to see we have given the =13.6 then we have assumed incompressible fluid constant, we are going to see how it works. So, this is point 1 here and this is point 2 here. (Refer Slide Time: 26:11)

In this chunk, we introduce the concept of barometers and their function in measuring atmospheric pressure. A barometer measures the pressure exerted by the atmosphere. The specific example given is relating to a height measurement (R) of 750 millimeters of mercury (Hg), which is a common standard used in pressure measurements. The value '13.6' refers to the density of mercury, indicating that it's being considered as a constant in this context.

Imagine you have a long tube filled with mercury, and it’s turned upside down into a shallow dish of mercury. The height of mercury that stays inside the tube is the atmospheric pressure at that moment. It's similar to how a straw works when you drink a beverage; the pressure difference helps keep the liquid in the straw while you sip.

So, we have seen this equation piezometric head is constant. So, we can also write let me erase this one. P1 is..... pressure at point one will be we have seen was R if you just go back here this P 2 - p 1 z2 - z 1 this is point 2. So, this = R. So, going to, can be written as a S * .. So, pressure will be P= S * R, on calculation it is going to give almost hundred 100,000 Pascal. So, this is one example as well of calculating the pressure p1 but this is also indicating how this barometer system works.

This chunk discusses how to calculate pressure at a specific point in a barometer using the piezometric head equation. The equation utilized compares pressure at two points (p1 and p2) including variations in elevation (z1 and z2). By understanding that the piezometric head is constant, we find that the atmospheric pressure can be expressed in terms of height (R) of mercury, ultimately calculating p1 to be approximately 100,000 Pascal.

Think of a water fountain where water rises and falls due to pressure. The same principle applies here; the height of mercury in the barometer is like how high the water would rise in that fountain, telling us how strong the atmospheric pressure is pushing down.

Now, we must also be a little aware about the pressure variations in a compressible fluid. So, there are 2 processes, one is perfect gas at constant temperature isothermal that we have been seeing till now. Secondly perfect gas with constant temperature gradient and actually you should try this at home. We are not going to cover this because this is the derivation is same as isothermal.

This section introduces the concept of pressure variations in compressible fluids, particularly distinguishing between two processes: isothermal and those with constant temperature gradients. It emphasizes that understanding pressure variation within compressible fluids is critical when designing systems that use gases, as they behave differently than incompressible fluids like liquids.

Consider a balloon. When you squeeze it, the air inside compresses and the pressure increases while maintaining the same temperature. If the balloon were to heat up while being squeezed, that would represent a different scenario where temperature affects pressure differently, leading to complex behaviors.

Now, we are going to do perfect gas at constant temperature that is isothermal process,... and if you integrate this, we are going to get because it is dp / p. So, that becomes ln(p2/p1) and dz is a simple integration yielding (Z2 - Z1). So, the pressure at point 2 in a constant temperature will be given as p = p1 e−[Mg/(R * Ts)g (Z2−Z1)].

In this chunk, we delve into how pressure in a perfect gas behaves during an isothermal process, where temperature remains constant. The pressure difference can be mathematically integrated to yield a specific formula. The derived equation helps to visualize how pressure changes relative to changes in elevation even if the temperature doesn’t vary.

Think of baking bread. The yeast we use generates gas that can expand, but if you keep the temperature steady while the dough rises, you can predict how much the gas will expand and how it will affect the volume and height of your bread.

Pressure measurement devices, we have discussed one already barometers, there are manometers a standard manometer or a differential manometer and the pressure transducers. This is out of the scope...

This part emphasizes the importance of pressure measurement devices such as barometers, manometers (both standard and differential), and pressure transducers. Each device has specific applications; for instance, barometers measure atmospheric pressure while manometers can measure pressures relative to atmosphere and pressure differences.

Picture a weather station, which uses barometers to provide the local atmospheric reading that helps predict the weather. Similarly, in hospitals, manometers can be used to monitor the pressure of gases that help in administering anesthesia to patients.

So, how high would the water rise in a manometer connected to a pipe containing water at 500 kPa?... pressure in a water distribution system, what will this height h in a manometer be?

In this example, the height of water in a manometer connected to a pipe is calculated given that the pressure in the system is 500 kPa. This helps illustrate how pressure values translate to physical heights in liquid. By applying the formula derived from pressure and density, students can visualize how different pressures would affect the height in a manometer.

Imagine measuring how high soda bubbles rise in a straw when you’re drinking. In this case, the pressure in the pipe pushes up the water in the manometer, similar to how carbonation pushes bubbles up into the straw when you sip.

--